PHP - Display Data Taken From Database Into Text Boxes

how i want to display data from database to look like this :



<table width="633" height="224" border="1">
  <tr bgcolor="#999900">
    <td width="45">Bil</td>
    <td width="121">Course_name</td>
    <td width="83">session</td>
    <td width="83">start_date</td>
    <td width="83">end_date</td>
    <td width="83">notes</td>
    <td width="89">pre-req</td>
  </tr>
  <tr bgcolor="#6A7AEA">
    <td rowspan="2">1.</td>
    <td rowspan="2" bgcolor="#6A7AEA">Math</td>
    <td>1st session </td>
    <td>1 jan 11 </td>
    <td>6 jan 11 </td>
    <td rowspan="2">&nbsp;</td>
    <td rowspan="2">&lt;image icon that will link to the oter site&gt; </td>
  </tr>
  <tr>
    <td bgcolor="#6A7AEA">2nd session </td>
    <td bgcolor="#6A7AEA">8 jan 11 </td>
    <td bgcolor="#6A7AEA">15 jan 11 </td>
  </tr>
  <tr>
    <td bgcolor="#0066CC">2.</td>
    <td bgcolor="#0066CC">English</td>
    <td bgcolor="#0066CC">1st session </td>
    <td bgcolor="#0066CC">1 feb 11 </td>
    <td bgcolor="#0066CC">6 feb 11 </td>
    <td bgcolor="#0066CC">&nbsp;</td>
    <td bgcolor="#0066CC">&lt;image icon that will link to the oter site&gt;</td>
  </tr>
  <tr>
    <td rowspan="2" bgcolor="#6A7AEA">3.</td>
    <td rowspan="2" bgcolor="#6A7AEA">Science</td>
    <td height="29" bgcolor="#6A7AEA">1st session </td>
    <td bgcolor="#6A7AEA">8 march 11 </td>
    <td bgcolor="#6A7AEA">15 march 11 </td>
    <td rowspan="2" bgcolor="#6A7AEA">&nbsp;</td>
    <td rowspan="2" bgcolor="#6A7AEA">&lt;image icon that will link to the oter site&gt;</td>
  </tr>
  <tr>
    <td bgcolor="#6A7AEA">2nd session</td>
    <td bgcolor="#6A7AEA">16 march 11 </td>
    <td bgcolor="#6A7AEA">21 march 11 </td>
  </tr>
</table>


** all the view data is called from database including the icon image
thanks...

This topic has been moved to JavaScript Help.

http://www.phpfreaks.com/forums/index.php?topic=321339.0

Hi guys,

Should be a simple 1.

If i have the following at the top of the page:


$page_views = $row['page_views'] + 1;
mysql_query("UPDATE table SET page_views='$page_views'");


and then the following at the bottom of the page:


echo $row['page_views'];


Should I see the page views as 1 the first time the page is visited, 2 the second time the page is visited, and so on......?
At the moment im seeing 0 on the first page visit, 1 on the second page visit, 2 on the third.....

I had this problem before on another page i was working, and i simply solved it by displaying the mysql query above the echo similar to the code above. However now it does not seem to be working.

Am i missing something really simple? lol

Thanks

This is actually a page to edit the drama details.
Database-  3 tables

1. drama                           dramaID    drama_title
                                              1              friends

2. drama_genre             drama_genreID     dramaID    genreID
                                               1                        1               2
                                               2                        1               1
                                               3                        1               3
3. genre                              genreID       genre
                                                1            comedy
                                                2            romance
                                                3            family
                                                4            suspense
                                                5            war
                                                6            horror

<?php       

$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$dbname = 'drama';
$link = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql');

$id = $_GET['dramaID'];

$link2 = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql');
$sql2 = "SELECT * from drama , drama_genre , genre
WHERE drama.dramaID='".$id."' AND drama_genre.dramaID='".$id."' AND drama.dramaID = drama_genre.dramaID AND drama_genre.genreID = genre.genreID";
$status2 = mysqli_query($link2,$sql2) or die (mysqli_error($link2));

$link3 = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql');
$sql3 = "SELECT * from genre ";
$status3 = mysqli_query($link3,$sql3) or die (mysqli_error($link3));

<form method='Post' action='doUpdate.php' enctype="multipart/form-data">

<?php while ($row2 = mysqli_fetch_assoc($status2)) { ?>
                  <td height="1"></td>
               
                             <td><select name="gdrop">      
                     
      <option value="<?php echo $row2['genreID'];?>"><?php echo $row2['genre'];?></option>
                     
      <?php while ($row3 = mysqli_fetch_assoc($status3)) { ?>
                        
      <option value="<?php echo $row3['genreID'];?>"><?php echo $row3['genre'];?></option>
                     
                     
      <?php }   mysqli_close($link3);  ?>
                     
 <input type='hidden' id='drama_genreID' name='drama_genreID' value = "<?php echo          $row2['drama_genreID']; ?>"/>
                 </select>

<input type="hidden" id="dramaID" name="dramaID" value = "<?php echo $row['dramaID'];?>"/>
<input type="submit" Value="Update"/>
 
The result was there were 3 dropdown menus but only the first dropdown menu has all 6 genres from my database and also the genre that belongs to the drama.

I'm also wondering how I can bring all the drama_genreIDs to my save(doupdate.php) page and update all 3 of them because it seems like only the last dropdown menu's data is saved.

And also how can I display only 6 genres instead of 7 with the genre that belongs to the drama , being set as the default selection.


                                       

As a complete newbie to php and webdesigning i have a following problem.I would like to retrieve the data from database and display it in a drop down menu.Then i should allow the user to select the values from drop down list along with other details,in other words i have to embed the drop down output as the form input for the user and store the form data in another table.I am running a xampp server and i am using php 5.4 version.Please help.My code is as follows.In this case project_name is displayed as the drop down output.but how do i use the same drop down output as a input in the form.
 
 
 
<html>
<head></head>
<body>
<?php
error_reporting(E_ALL ^ E_DEPRECATED);
include 'connect.php' ;
$tbl_name="projects";


$sql="SELECT project_name FROM $tbl_name ";

$result=mysql_query($sql);


if($result === FALSE) {
    die(mysql_error());
}
?>


<form name="resources" action="hourssubmit.php"  method="post" >

<?php
echo "<select name='project_name'>";
while ($row = mysql_fetch_array($result)) {
    
    echo "<option value='" . $row['project_name'] ."'>" . $row['project_name'] ."</option>";
}
echo "</select>";
?>

</form>
</body>
</html>

i have 8 division (div), i want to display 4 rows in 4 division and the remain 4 rows in the next 4 division here is my code structure for carousel

<div class="nyie-outer">    
<div class="container">
    <div  id="nyieFull" class="carousel slide carousel-multi-item" data-interval="false" data-ride="carousel">
        <div class="outer-controls-top">
        <div class="nyie-title"><h5>THIS IS NYIE</h5></div>
        <div class="controls-top">    
            <a class="btn-floating" href="#nyieFull" data-slide="prev"><i class="fa fa-chevron-left"></i></a>
            <a class="btn-floating" href="#nyieFull" data-slide="next"><i class="fa fa-chevron-right"></i></a>
        </div>
        </div>
        <div class="carousel-inner" role="listbox">
            <div class="carousel-item active">
                 first row

                 second row

                 third row

                 fourth row
            </div><!--/.First four rows here-->
            <div class="carousel-item">

                  fifth row

                 sixth row

                 seven throw

                 eighth row

            </div><!--/.second four rows here-->
        </div
    </div>    
</div> 
</div>

 

 

sql code   

CREATE TABLE  product(
prodId INT(11) NOT NULL UNIQUE AUTO_INCREMENT, 
image VARCHAR(255) NOT NULL,
prodName VARCHAR(255) NOT NULL,
prodPrice VARCHAR(255) NOT NULL,
description TEXT NOT NULL
PRIMARY KEY (prodId) 
);

php code

<?php 
include_once('db.php');
$sql = "SELECT image,prodName, prodPrice, description FROM product ORDER BY prodId ASC LIMIT 4";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
    // output data of each row
        while($row = mysqli_fetch_array($result)) {
            $activeprod =  '
            <div class="col-lg-3">
            <a href="more.php" class="newAdded">    
            <figure class="figure">   
               <img src="images/'. $row['image'] .'" alt="" class="img-responsive" >
                <figcaption class="figure-caption">
                    <span class="prodName"><b>Name: </b>'. $row['prodName'] .'</span>
                    <span class="prodPrice"><b>price: </b>'. $row['prodPrice'] .'</span>
                    <span class="desc"><b>description: </b>'. $row['description'] .'</span>    
                </figcaption>    
            </figure> 
            </a>
            </div>
            ';
    }
} 
?>

how can i echo that result in those rows 

 

 

 

I currently am working on a project where I code a "simple" telephone directory.

There are three main tasks that it needs to do:
1. Directory.php(index page) has a "First Name" and "Last Name" field and a search button. When a name is searched from the directory.txt file, it displays First Name, Last Name, Address, City, State, Zip and phone in findinfo.php in designated text boxes...first name, last name, etc.

2. From the findinfo.php, like previously stated, the users information is listed in the appropriate text boxes. From there, there is an update button that will overwrite the user's information to directory.txt if that button is selected. It will then say the write was sucessful.

3. (completed this step) From the index page, there is a link that will take you to addnew.php where you enter First Name, Last Name, Address, City, State, Zip and phone in a web form and write it to directory.txt.

 This is the php code for the third step:

<?php
$newentryfile = fopen("directory.txt", "a+");
$firstname = $_POST['fname'];
$lastname = $_POST['lname'];
$address = $_POST['address'];
$city = $_POST['city'];
$state = $_POST['state'];
$zip = $_POST['zip'];
$phone = $_POST['phone'];
$newentry = "$firstname $lastname\n\r $address\n\r $city, $state $zip\n\r $phone\n\r";
if (flock($newentryfile, LOCK_EX)) {
if (fwrite($newentryfile, $newentry) > 0)
echo "<p>" . stripslashes($firstname) . " " .
 stripslashes($lastname) . " has been added to the directory.</p>";
else
echo "<p>Registration error!</p>";
flock($newentryfile, LOCK_UN);
}
else
echo "<p>Cannot write to the file. Please try again later</p>";
fclose($newentryfile);

if(empty($firstname) || empty($lastname) || empty($address) || empty($city) || empty ($state) || empty($zip) || empty($phone))
{
echo "<p>Please go back and fill out all fields.</p>";
}
?>


So to sum it all up, what would be my best approach? I am totally stumped and not sure which function to use. Should I work my way from step 1 to step 2? I see it as when I do the search for the name from directory.php, it takes me to findinfo.php, listing the users information in the text boxes. From there, if I needed to, having the user's information already listed I could hit the update button to overwrite the new information to directory.txt. Doing the update when then tell me that the write was successful.

I have literally been scouring the internet for hours. What would be the best function to do this? I hope I was clear enough. Please help me out and thank you for your time.

I have a set of 26 checkboxes that are stored in an array. The below code dumps the comma separated array values into a single record column.


Code: [Select]
// dumping the type of job checkboxes

$type = $_POST['type'];
  var_dump($type);
 
    // Setting up a blank variable to be used in the coming loop.
  $allStyles = "";
 
  // For every checkbox value sent to the form.
  foreach ($type as $style) {
    // Append the string with the current array element, and then add a comma and a space at the end.
    $allTypes .= $style . ", ";
  }
 
    // Delete the last two characters from the string.
  $allTypes = substr($allTypes, 0, -2);


// end dumping the type of job checkboxes


This works great! But my needs go past this. The record can be edited. So I need to get the values back out of the database and populate the correct check boxes for editing.

The checkboxes are generated by this code.


Code: [Select]
<div class="text">Job Type&nbsp;(<a class="editlist" href="javascript:editlist('listedit.php?edit=typelist',700,400);">edit list</a>):</div>
<div class="field">

<?PHP

$connection=mysql_connect ("localhost", "name", "pass") or die ("I cannot connect to the database.");
$db=mysql_select_db ("database", $connection) or die (mysql_error());
$query = "SELECT type FROM typelist ORDER BY type ASC";
$sql_result = mysql_query($query, $connection) or die (mysql_error());
$i=1;
echo '<table valign="top"><tr><td>';
while ($row = mysql_fetch_array($sql_result)) {
$type = $row["type"];
    if ($i > 1 && $i % 26 == 0)
      echo '</td><td>';
  else if ($i)
  echo '';
  ++$i;
  echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type'><span style='color:#000;'>$type</span></input><br/>";
}
  echo '</td></tr></table>';

?>
 
</div>


I am having a problem getting these checkboxes repopulated. I think that a way to do this is to:

Get the values out of the database in a array
Compare the values the list of all possible values.
Check the boxes of the values confirmed to be there.

I didn't build the application so I need help to work through this problem.

hello.

i have a question for you.

say i have a group of tick boxes and i select a couple.

would i put them in to the database as varchar ?
like this:

id   varchar
1    1, 2, 3

or

id   varchar
1    '1', '2', '3'


then my next question is how would i use them.
i guess i would explode them or something.

the thing is, i have some code that currently pull 1 item from the database (using varchar) and echos it fine but when i add more i get nothing back.

thanks
rick

Hi, guys.
I am a php newb and stuck (see title).
I it will only send the value of 1 tick box over to the page "delete_message.php"

This is what i have so far:

Code: [Select]
<?php
// get messages
$result = mysql_query("SELECT * FROM tbl_pvt_msg WHERE to_UID='$UID'");
if (!$result) die("Query to show fields from table failed");
while ($data=mysql_fetch_array($result)) {

$_SESSION['from_UID']=$data['1'];
UIDtoemail();
$from_email=$_SESSION['from_email'];
$contents=$data['3'];
$timesent=$data['4'];
$msg_ID=$data['0'];
echo "<form action='scripts/delete_message.php' method='POST' >";
echo "<tr><td>$from_email</td><td>$contents</td><td>$timesent</td><td><input type='checkbox' value='$msg_ID' name='msg_ID' /></td> </tr>";
}
// turn uid into email
function UIDtoemail(){
$from_UID=$_SESSION['from_UID'];
$result = mysql_query("SELECT * FROM tbl_usr WHERE UID='$from_UID'");
if (!$result) die("Query to show fields from table failed");
$data=mysql_fetch_array($result);
$_SESSION['from_email']=$data['7'];

}
?>
</table>
<input type="submit" value="Delete" id="delete" />
</form>

Hi,

I had a form which contains, 8 select boxes. In those 8 select boxes, 3 of the boxes are interlinked, like, I am able to generate data in other 2 select boxes, if one box is been selected. i.e. based upon selection of one select box, and i am able to generate data in other select box. So, 3 select boxes are interlinked. And remaining 5 selects are optional, if the user selects, then we need to generate data based upon selection of the user.

Here the main issue is how to grab data in the post back form, like, how to capture the number of selects boxes selected and what are those selected, and based upon that we need to generate data.

Here one more important thing is that, in the report generation form(postback form), I need to show the data in a table. So , in table I had 9 columns and these would be common to whatever the selection made, but, only the requeired data would be changed based upon the selection of the remaining 5 optional boxes.

Hope you got me!!!!

Hello to all,

I have problem figuring out how to properly display data fetched from MySQL database in a HTML table.

In the below example I am using two while loops, where the second one is nested inside first one, that check two different expressions fetching data from tables found in a MySQL database. The second expression compares the two tables IDs and after their match it displays the email of the account holder in each column in the HTML table.

The main problem is that the 'email' row is displayed properly while its while expression is not nested and alone(meaning the other data is omitted or commented out), but either nested or neighbored to the first while loop, it is displayed horizontally and the other data ('validity', 'valid_from', 'valid_to') is not displayed.'

Can someone help me on this, I guess the problem lies in the while loop?
Here is part of the HTML code:

<thead>
 <tr>
 <th data-column-id="id" data-type="numeric">ID</th>
 <th data-column-id="email">Subscriber's Email</th>
 <th data-column-id="validity">Validity</th>
<th data-column-id="valid_from">Valid From</th>
 <th data-column-id="valid_to">Valid To</th>
</tr>
</thead>

Here is part of the PHP code:

 

    <?php
    while($row = $stmt->fetch(PDO::FETCH_ASSOC))
    {
        echo '
            <tr>
                <td>'.$row["id"].'</td>
        ';
        while ($row1 = $stmt1->fetch(PDO::FETCH_ASSOC)) {
        echo '
                <td>'.$row1["email"].'</td>
        ';
        }
        if($row["validity"] == 1) {
            echo '<td>'.$row["validity"].' month</td>';
        }else{
            echo '<td>'.$row["validity"].' months</td>';
        }
        echo '  <td>'.$row["valid_from"].'</td>
                <td>'.$row["valid_to"].'</td>
        </tr>';
    }
    ?>

P.S. Please know that I have purposely omitted the connection part and the methods in the class that are used in this code to keep the spot of the problem as clean as possible. If you need me to rephrase the question, please ask.

Thank you.

Hi, I've currently started to modify a chat script of mine to output a moderation panel but the moderation page seems empty(blank) every time I load it. What im trying to do is to take the ID part in my URL via the $_GET and look it up in my database table in the column named id, then select that specific row to be able to retrieve the StringyChat_ip and place it into another table to ban the IP and the second thing im trying to do is to be able to delete the specific row from my table.   My Http link look something like    http://imagecrab.fre.../ban.php?id=159   and my ban.php page where I want to lookup the 159 part and do the banning etc looks like

<?
include("admin_code_header.php");


if ($_POST["DeletePost"]) {
    $id = $_POST['id'];
    $query = "DELETE FROM ".$dbTable." WHERE id='".$id."'"; 
    mysql_query($query);
    echo "ID removed from system: ".$id;
  }


if ($_POST["BanIP"]) {
    $IP_To_Add = $_POST["ip"];
    
      $sql = "INSERT INTO ".$IPBanTable." (ip) VALUES (\"$IP_To_Add\")";
      $result = mysql_query($sql);
  }
$result = mysql_query("SELECT * FROM ".$dbTable." WHERE id='".$id."'",$db);

  while ($myrow = mysql_fetch_array($result)) {
    $msg = $myrow["StringyChat_message"];
$idm = $myrow["id"];



?>
<html>
<form name="form<? echo $myrow["id"];?>" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">

<input name="DeletePost" type="submit" id="DeletePost" value="Delete">

<input name="BanIP" type="submit" id="BanIP" value="Ban <? echo $myrow["StringyChat_ip"];?>">
    </form>
</html>
<?
}
?>

hi guys, im new to this forum

I'm new also to php, I need help from you guys:

I want to display personal information from a certain person (the data is on the mysql database) using his name as a link:

example:
               (index.php)
               names
                          1. Bill Gates
                          2. Mr. nice Guy
           
i want to click Bill Gates
 
   (output.php)
     Name: Bill Gates
     Country:xxxx
     Age: xx   
                 etc.

How can i make this or how to learn this?